Does interpolation with vanishing degree slack $(1+\epsilon(n))n$ still force a.e. divergence? (Erdős #1152)

For each $n\geq 1$ fix $n$ distinct nodes $x_{1n},\ldots,x_{nn}\in [-1,1]$, and let $\epsilon=\epsilon(n)\to 0$. Does there always exist a continuous function $f:[-1,1]\to \mathbb{R}$ such that, whenever $p_n$ is a sequence of polynomials with degrees $\deg p_n<(1+\epsilon(n))n$ satisfying the interpolation conditions $p_n(x_{kn})=f(x_{kn})$ for all $1\leq k\leq n$, the sequence fails to converge pointwise: $p_n(x)\not\to f(x)$ for almost all $x\in [-1,1]$? (Note that with slack the interpolant is not unique, so $f$ must defeat EVERY admissible sequence $p_n$.)

Acceptance. FULLY RESOLVES: a complete proof (machine-checkable Lean/Coq preferred, else a full written proof) that for every node array and every $\epsilon(n)\to 0$ such an $f$ exists; OR a counterexample: an explicit node array and an explicit slack function $\epsilon(n)\to 0$, together with a proof that for EVERY continuous $f$ there is an admissible interpolating sequence $p_n$ (degrees $<(1+\epsilon(n))n$) with $p_n(x)\to f(x)$ on a set of positive measure. ADVANCES: (a) resolving the question for a specific natural node array (e.g. Chebyshev nodes) with vanishing slack, in either direction, with proof; (b) a quantitative threshold theorem identifying growth conditions on $\epsilon(n)$ (e.g. $\epsilon(n)\gg 1/\log n$ vs $\epsilon(n)\ll 1/\log n$) that separate the fixed-slack convergence regime of [EKS89] from the divergence regime, with proof; (c) the weaker adversarial conclusion — an $f$ forcing $p_n(x)\not\to f(x)$ on a set of positive measure (rather than almost everywhere) for every admissible sequence — proved for all arrays. Deliver the proof file(s).